The conditioning number of a helix, part 2

Recall the previous attempt to find the conditioning number of a helix (see post "The conditioning number of a helix, part 1," 2016-10-31). Here we complete the approach and although exact solutions are hard to find, we give close approximations.

The setting was a helix $C$ of radius $r$ and stretch $c$, so given as the zero locus of $x-r\cos(z/c)$ and $y-r\sin(z/c)$, and we wanted to find where the normal plane at a point $p\in C$ intersects $C$ again. It may intersect $C$ several times, but we are only interested in the shortest distances. Without loss of generality, assume that $p=(r,0,0)$. The normal plane at $p$ is then given by
\[
0 =
\det\begin{bmatrix}
x-r\cos(p_z/c) & y-r\sin(p_z/c) & z-p_z \\
1 & 0 & r\sin(p_z/c)/c \\
0 & 1 & -r\cos(p_z/c)/c
\end{bmatrix}
=
\det\begin{bmatrix}
x-r & y & z \\
1 & 0 & 0 \\
0 & 1 & -r/c
\end{bmatrix}
=
\frac rc y + z.\]Since the cylinder on which the helix $C$ lies is $x^2+y^2=r^2$, the curve $C'$ representing the intersection of the plane with the cylinder is given by the zero locus of $\pm r\sqrt{x^2-r^2} + cz$. This allows us to find the intersection with the helix. However, since $C$ is parametrized with $z$ the free variable and $C'$ with $x$ free, its is easier to switch to cylindrical coordinates
\[
\left(r = \sqrt{x^2+y^2},\theta = \arctan(y/x), z = z\right).\]Doing so gives a nice description of $C$ and $C'$ as below.\[
\begin{array}{r l c l}
C : &  (r\cos(z/c),r\sin(z/c),z) & = & (r,\theta,\theta c) \\
C' : &  (x,-\sqrt{r^2-x^2},r\sqrt{r^2-x^2}/c) & = & (r.\theta,r^2\sin(\theta)/c)
\end{array}\]The switch in coordinates is represented by the diagram below, where we have only used the top half of $C'$.

Finding $C\cap C'$ is equivalent to solving $\frac{c^2}{r^2} = \frac{\sin(\theta)}\theta$ for $\theta$, a task that can not be solved exactly. Instead we take the tangent lines to $C'$ on the unrolled cylinder at its base, and see where those intersect the line $\theta c$. Inspecting the areas of the tangent lines closer and calculating the euclidean distances in $\R^3$ from $p$ to $a$ and $b$, which is, I can't believe I'm saying this, a great exercise for the reader, we get the distances to be
\[
d(p,a) = \sqrt{2r^2\left(1+\cos\left(\frac{\pi c^2}{r^2-c^2}\right)\right) + \left(\frac{\pi cr^2}{r^2-c^2}\right)^2},
\hspace{.2cm}
d(p,b) = \sqrt{2r^2\left(1-\cos\left(\frac{2\pi c^2}{r^2+c^2}\right)\right) + \left(\frac{2\pi c r^2}{r^2+c^2}\right)^2}.\]Truthfully, the diagrams are tricky to draw in TikZ and I don't want to simply have a scan of some rough work. More importantly, $d(p,a) = d(p,b)$ implies $c = r/\sqrt 3$, meaning that when the stretch $c$ is larger than $r/\sqrt 3$, the normal planes certainly do not intersect the helix again.